Question

Derivative of sin^mx×cos^nx

Answer 1

we know

y = f(x).g(x) then ,

Dy/dx = f(x).DG(x)/DX+ g(x).DF(x)/DX

y = sin^m x.cosⁿx

differentiate with respect to x

dy/dx = sin^mx . {n.cos^(n-1)x (-sinx)} + cosⁿx { m.sin^(m -1)x (cosx) }

= - n.sin^(m+1)x .cos^(n-1)x +m.cos^(n+1)x.sin^(m-1)x